Find the limit of the sequence $y_n = nleft(npi + fracpi2 - x_nright)$ [on hold]$(y_n)$ bounded sequence and $limfrac(x_n)(y_n)=0$ show $lim(x_n) = 0$Find the limit of the sequence with $n$th term $x_n = left(frac2n+3n^2right)^n$Convergence and upper bound of $sum_n fracsinleft[left(pi-frac14right)(x_n-x_m)right]x_n-x_msinleft(fracx_n+x_m4right)$.If $x_nto 0$ and $y_n=fracx_1-x_2-…-x_nn$ then $y_nto 0$Find $limlimits_n to infty fracx_nn$ when $limlimits_n to infty x_n+k-x_n$ existsProve that the sequence $z_n$, where $z_n:= x_n-y_n$, converges and $lim (x_n-y_n)=lim z_n = (lim x_n) - (lim y_n)$sequence $x_n+1=x_n+e^-2018x_n$Asymptotics of the sequence defined by $x_n+1 = x_n + frac2x_n$, $x_0=1$Limit of the infinite sequence involving $a_n, s_n$Concluding whether $(y_n)_n$ is a bounded sequence

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Find the limit of the sequence $y_n = nleft(npi + fracpi2 - x_nright)$ [on hold]


$(y_n)$ bounded sequence and $limfrac(x_n)(y_n)=0$ show $lim(x_n) = 0$Find the limit of the sequence with $n$th term $x_n = left(frac2n+3n^2right)^n$Convergence and upper bound of $sum_n fracsinleft[left(pi-frac14right)(x_n-x_m)right]x_n-x_msinleft(fracx_n+x_m4right)$.If $x_nto 0$ and $y_n=fracx_1-x_2-…-x_nn$ then $y_nto 0$Find $limlimits_n to infty fracx_nn$ when $limlimits_n to infty x_n+k-x_n$ existsProve that the sequence $z_n$, where $z_n:= x_n-y_n$, converges and $lim (x_n-y_n)=lim z_n = (lim x_n) - (lim y_n)$sequence $x_n+1=x_n+e^-2018x_n$Asymptotics of the sequence defined by $x_n+1 = x_n + frac2x_n$, $x_0=1$Limit of the infinite sequence involving $a_n, s_n$Concluding whether $(y_n)_n$ is a bounded sequence













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$begingroup$


If $x_n$ is the solution in the interval $left(n pi, n pi + fracpi2right)$ of the equation $ tanx=x$, where $n$ is a natural number, find the limit of the sequence $y_n = nleft(npi + fracpi2 - x_nright)$ when $n$ goes to $infty$. I tried to apply Cesaro Stolz or Taylor series but doesn't work. Can you give me some hints?










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$endgroup$



put on hold as off-topic by user21820, Xander Henderson, RRL, Saad, Parcly Taxel 2 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Xander Henderson, RRL, Saad, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    The $n$th root of $tan x = x$ is $x_n = pi n + pi/2 - epsilon$, and you need $tan x_n$ to approximately match $pi n$. From this you can find $epsilon$ to the first order. Incidentally, $x_n$ is the $n$th positive zero of the Bessel function $J_3/2(x)$.
    $endgroup$
    – Maxim
    yesterday















0












$begingroup$


If $x_n$ is the solution in the interval $left(n pi, n pi + fracpi2right)$ of the equation $ tanx=x$, where $n$ is a natural number, find the limit of the sequence $y_n = nleft(npi + fracpi2 - x_nright)$ when $n$ goes to $infty$. I tried to apply Cesaro Stolz or Taylor series but doesn't work. Can you give me some hints?










share|cite|improve this question











$endgroup$



put on hold as off-topic by user21820, Xander Henderson, RRL, Saad, Parcly Taxel 2 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Xander Henderson, RRL, Saad, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    The $n$th root of $tan x = x$ is $x_n = pi n + pi/2 - epsilon$, and you need $tan x_n$ to approximately match $pi n$. From this you can find $epsilon$ to the first order. Incidentally, $x_n$ is the $n$th positive zero of the Bessel function $J_3/2(x)$.
    $endgroup$
    – Maxim
    yesterday













0












0








0





$begingroup$


If $x_n$ is the solution in the interval $left(n pi, n pi + fracpi2right)$ of the equation $ tanx=x$, where $n$ is a natural number, find the limit of the sequence $y_n = nleft(npi + fracpi2 - x_nright)$ when $n$ goes to $infty$. I tried to apply Cesaro Stolz or Taylor series but doesn't work. Can you give me some hints?










share|cite|improve this question











$endgroup$




If $x_n$ is the solution in the interval $left(n pi, n pi + fracpi2right)$ of the equation $ tanx=x$, where $n$ is a natural number, find the limit of the sequence $y_n = nleft(npi + fracpi2 - x_nright)$ when $n$ goes to $infty$. I tried to apply Cesaro Stolz or Taylor series but doesn't work. Can you give me some hints?







sequences-and-series






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









rtybase

11.4k31533




11.4k31533










asked yesterday









rafarafa

618212




618212




put on hold as off-topic by user21820, Xander Henderson, RRL, Saad, Parcly Taxel 2 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Xander Henderson, RRL, Saad, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.







put on hold as off-topic by user21820, Xander Henderson, RRL, Saad, Parcly Taxel 2 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Xander Henderson, RRL, Saad, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.











  • $begingroup$
    The $n$th root of $tan x = x$ is $x_n = pi n + pi/2 - epsilon$, and you need $tan x_n$ to approximately match $pi n$. From this you can find $epsilon$ to the first order. Incidentally, $x_n$ is the $n$th positive zero of the Bessel function $J_3/2(x)$.
    $endgroup$
    – Maxim
    yesterday
















  • $begingroup$
    The $n$th root of $tan x = x$ is $x_n = pi n + pi/2 - epsilon$, and you need $tan x_n$ to approximately match $pi n$. From this you can find $epsilon$ to the first order. Incidentally, $x_n$ is the $n$th positive zero of the Bessel function $J_3/2(x)$.
    $endgroup$
    – Maxim
    yesterday















$begingroup$
The $n$th root of $tan x = x$ is $x_n = pi n + pi/2 - epsilon$, and you need $tan x_n$ to approximately match $pi n$. From this you can find $epsilon$ to the first order. Incidentally, $x_n$ is the $n$th positive zero of the Bessel function $J_3/2(x)$.
$endgroup$
– Maxim
yesterday




$begingroup$
The $n$th root of $tan x = x$ is $x_n = pi n + pi/2 - epsilon$, and you need $tan x_n$ to approximately match $pi n$. From this you can find $epsilon$ to the first order. Incidentally, $x_n$ is the $n$th positive zero of the Bessel function $J_3/2(x)$.
$endgroup$
– Maxim
yesterday










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